# Schwarzschild metric system of geodesic differential equations

Suppose that we are given the Schwarzschild metric and its Lagrangian $$L=-(1-\frac{R}{r})t'^2 + (1-\frac{R}{r})^{-1}r'^2+r^2 \theta'^2+r^2 \sin^2(\theta)\phi'^2$$ where R=$$r_s=2GM$$ and $$x'=\frac{d}{d \tau}$$ $$\forall x \in A:A={t',\theta',r',\phi'}$$ The set of geodesic equations given by the Schwarzschild metric are $$2(1-\frac{r_s}{r})t''=0$$ $$2r\theta'^2-\frac{r_st'^2}{r^2}-\frac{r_sr'^2}{(r-r_s)^2}-\frac{2rr''}{r-r_s}=0$$ $$\phi'^2\sin(2\theta)-2r^2\theta''=0$$ $$-2\phi''^2\sin^2(\theta)=0.$$ To solve for these geodesics we need to consider a transformation of differential equation order. We need to change this system of 4 second order ODEs into a system of 8 first order ODEs. Question is how would I accomplish this? I know how to transform single second order ODEs like $$y''-4y'-2y=0$$ into a system of 2 first order ODEs. How would I go about doing this for a system,not individual ODEs?

First, it is most definitely not necessary to transform your second order differential equations into first order differential equations in order to solve them. However, if you wish to do so for whatever reason, then the systematic way to do that is by using the Hamiltonian approach.

Here we define the momenta conjugate to each variable as $$p_i = \partial \mathcal{ L}/\partial \dot q_i$$. This gives us a set of equations which we can solve for the $$\dot q_i$$ in terms of the $$p_i$$. Then we calculate the Hamiltonian as $$H = \Sigma p_i \dot q_i -\mathcal{L}$$, where we substitute the above expressions for $$\dot q_i$$ so that we have a function of the $$p_i$$ and the $$q_i$$ with no $$\dot q_i$$ terms remaining.

Once we have done that, we can get a system of first order differential equations by solving Hamilton's equations: $$\frac{dq_i}{d\tau}=\frac{\partial H}{\partial p_i}$$ $$\frac{dp_i}{d\tau}=-\frac{\partial H}{\partial q_i}$$

Here this gives us: $$\begin{array}{c} \dot t=\frac{r p_t}{2 (R-r)} \\ \dot p_t=0 \\ \dot r=\frac{1}{2} p_r \left(1-\frac{R}{r}\right) \\ \dot p_r=\frac{1}{4} \left(\frac{2 \left(\csc ^2(\theta ) p_{\phi }^2+p_{\theta }^2\right)}{r^3}-\frac{R p_r^2}{r^2}-\frac{p_t^2}{R-r}-\frac{r p_t^2}{(R-r)^2}\right) \\ \dot \theta =\frac{p_{\theta }}{2 r^2} \\ \dot p_{\theta }=\frac{\cot (\theta ) \csc ^2(\theta ) p_{\phi }^2}{2 r^2} \\ \dot \phi =\frac{\csc ^2(\theta ) p_{\phi }}{2 r^2} \\ \dot p_{\phi }=0 \\ \end{array}$$

Notice that $$\dot p_t = 0$$ gives us a conserved energy $$p_t=E$$ and $$\dot p_\phi=0$$ gives us a conserved angular momentum $$p_\phi = L$$ which we can substitute into the remaining six equations to get a system of six first order differential equations.

• Thank you for clearing things up. For finding the H term, do we find the expression for H from the metric we are given or from the Lagrangian we get from the metric?
– aygx
Nov 8, 2021 at 2:25
• @aygx the Hamiltonian comes from the Lagrangian.
– Dale
Nov 8, 2021 at 2:54
• Thats what I assumed, thank you. The hamiltonian approach is rather straight foward, however, Is there a method to solve the system of equations in the form of the euler lengrange equations or is the Hamiltonian approach the only solution?
– aygx
Nov 8, 2021 at 11:34
• @aygx Yes, you can solve second order differential equations directly. There are both analytical methods as well as numerical methods for directly solving such equations. A full exposition on those methods is well beyond the scope of a question or a comment here, but that should be covered in any course on differential equations. Typically something that you would learn in the sophomore year of an engineering curriculum.
– Dale
Nov 8, 2021 at 17:12